## PHYS345 Electricity and Electronics

**Quiz 4:** ac Circuit Analysis

Find the voltage across the capacitor.

**Evaluation of the Element Reactances:**

**Circuit Reduction:**

This is a simple circuit, meaning that the network of passive elements may be reduced to a single
equivalent impedance. The progression to the simple circuit is indicated below.

**Evaluation of Source Current:**

The equivalent impedance is *Z*_{eq} = (9+j9) ohm = 12.7 ohm \/ -45 deg.

The source current is *i*_{s} = v_{s} /*Z*_{eq} = (36 V) / (12.7 ohm \/ -45) = 2.83 A \/ -45 deg.

**Evaluation of the Voltage across the Element:**

Note that at the node, the current splits equally since the two parallel branches have the same impedance.
Thus the current through the capacitor is *i*_{c} = 1.42 A \/ -45 deg.
Consequently the voltage across the capacitor is v_{c} = *i*_{c}*Z*_{c} = (1.42 A \/ -45)(6 ohm \/ -90) = 8.52 V \/ -135 deg.

"http://www.physics.udel.edu/~watson/phys345/quiz/04soln.html"

Last updated Oct. 1, 1999.

Copyright George Watson, Univ. of Delaware, 1999.